Discrete Mathematics
Summer 03
Assignment # 3 : Solutions
Section 3.1
12.“The sum of any two odd integers is even”.
Given: Two odd numbers a and b.
Prove: a+b is even.
From definition of odd number, a=2n+1 and b=2k+1 for integers n, k.
a+b = 2n+1+2k+1 = 2n + 2k+ 2 = 2*(n+k+1)
Let integer s = n+k+1.
Then we have a+b = 2*s.
Therefore, from definition of even, a+b is even. 
24.The problem with the given proof is that it “begs the question.”
Explanation: Proof assumes what it is to be proved (that m*n is even and that it can be
expressed as 2r). And later it assumes this to be true ((2p)*(2q+1) = 2r)
34.Statement: For all integers m, if m > 2 then m2-4 is composite.
This statement is false. A counter example is m=3.
Although 3 > 2, 32 - 4= 5, which is prime.
Section 3.2
17. Statement: Given any two distinct rational numbers r and s with r < s, there’s a rational
number x such that r < x < s.
Given: Any two rational numbers r and s, with r < s.
Prove: There exists a rational x s.t. r < x <s.
From definition of rational, let r = a/b, and s = c/d for integers a, b, c, d, with b≠0
and d≠0.
Re-express the fractions r and s over the same denominator 2bd:
r = 2da/2bd and s=2bc/2bd.
Since r<s, 2da<2bc.
Then (2da+2bc)/2 = da+bc is the odd integer that lies halfway between the integers 2da
and 2bc.
Hence (da+bc)/2bd is a rational number that lies between r and s.
Discrete Mathematics
Summer 03
Problem A.
Prove 0.9999… = 1.0
Let x = 0.9999…
Then 10x = 9.9999…
Subtract:
10x = 9.9999…
-x = 0.9999…
-------------------------9x = 9.0
Then x = 1.0 
Section 3.3
24. Statement: For all integers a, b, and c, if a | bc then a | b or a | c.
This statement is false. A counter example is a=12, b=3, c=4.
12 | 3*4, but neither 12 | 3 nor 12 | 4.
33. Suppose n is an integer such that 2*3*4*5*n = 29*28*27*26*25.
Does 29 | n? Why?
Yes. Since 29 is one of the prime factors of the right hand side of the equation, it is also a
prime factor of the left-hand side (by the unique factorization theorem). But 29 does not
equal a prime factor of 2, 3, 4, or 5 (because it’s too large). Hence 29 must occur as one
of the prime factors of n, and so 29 | n.
Section 3.4
26. Proposition: For any integer n 2, n2-3 is never divisible by 4.
Proof:
If n is even, then n2 is even, and so n2-3 is odd and hence not divisible by 4. Thus the
remaining case to consider is when n is odd.
Any integer n can be put into one of the four cases 4q, 4q+1, 4q+2, and 4q+3. Since
4q and 4q+2 are even, only the cases 4q+1 and 4q+3 need be considered.
Case 1: n
= 4q+1.
n2-3 = 16q2+1+ 8q –3
= 16q2+8q-2.
Not divisible by 4. (Although the first two terms are, last term is not)
Discrete Mathematics
Summer 03
Case 2: n
= 4q+3.
2
n -3 = 16q2+9+24q–3
= 16q2+24q+6.
Not divisible by 4. (Although the first two terms are, last term is not)
Therefore, we have proved that n2-3 is never divisible by 4, for any integer n2. 
30. Statement: If n is an odd integer then n4 mod 16 = 1.
Given: An odd integer n.
Prove: n4 mod 16 = 1.
n4 mod 16 = 1 is equivalent to saying (n4-1)mod 16 = 0.
So, we want to prove that n4-1 is a multiple of 16.
Observe that n4 –1 = (n2-1)(n2+1)
[Note: Whenever you see a a2-1 be aware of the handy fact that a2-1 = (a-1)(a+1).]
Since n is odd, it can be written in one of two forms: 4q+1, 4q+3.
In the first case, n2 = 16q2 + 8q +1, and
in the second case n2 = 16q2 + 24q + 9.
In either case, observe that (n2 –1) is a multiple of 8 and (n2+1) is a multiple of 2.
Hence their product, which equals (n4-1), is a multiple of 16. 
Section 3.5
20. Statement: For all real numbers x and y, xy = x. y
This statement is false. A counter example is x=1.5, y=2.0.
Then x = 2, y = 2, and xy = 3.0 = 3. And 2*2  3.
 n 2   n  1  n  1 
28. Statement: For any odd integer n,    

.
 4   2  2 
 n2  1
 n  1  n  1 


equals



4
 2  2 


Discrete Mathematics
Summer 03
 n2  n2  1
Therefore, we want to prove:   
.
4
4
From definition of odd we let n= 2k+1.
n 2  1 4 k 2  4k  1  1
Then the right hand side equals:

= k2  k .
4
4
2
2
 n   4k  4 k  1 
And the left hand side equals:    
 k 2  k  0.25  k 2  k .

4
4 

Therefore, the right hand side equals the left hand side. 
Section 3.6
4. Statement: There’s no least positive rational number.
Proof:
Suppose not. Suppose there exists a least positive rational number, call it R.
This means that Rr for all rational numbers r. Say R = a/b, for integers a, b, b0.
Now we multiply b by 2 to get: R’ = a/2b. R’ is still a positive rational number and in fact
R’<R since they have the same numerator and R’ has a larger denominator. But then this
means that R’ is smaller than the least positive rational number. Therefore, it’s a
contradiction. 
22. Statement: If a and b are rational numbers, b 0, and r is an irrational number, then
a+br is irrational.
Proof:
Suppose a+br is rational. 
br is rational 
br/r = r is rational.
------- CONTRADICTION
Section 3.7
10. Statement: 32 is irrational.
Proof:
Suppose not. Then there are integers m and n with no common factors, such that
3
2=m/n. Taking the cube of both sides gives: 2 = m3/n3 or 2n3 = m3. This means that m3
is even.
Lemma1: odd*odd = odd.
Proof: (2m+1)*(2n+1) = 4mn+2m+2n+1 = 2(2mn+m+n) +1.
Discrete Mathematics
Summer 03
Lemma2: If m3 is even, then m is even.
Proof: We show the contrapositive, namely, that m odd  m3 odd.
Suppose m is odd.
Then m3 = odd*odd*odd = (odd*odd) * odd = odd*odd = odd. (By Lemma 1)
Hence m is even. Say m=2*k for some integer k. Then m3 = (2k)3 = 8k3.
From before, m3 = 2n3. So 8k3 = 2n3, and thus 4k3 = n3. Hence n3 is even, and so by
Lemma 2, n is even.
But the fact that both m and n are even contradicts our earlier statement that they have no
common factors. This contradiction completes the proof.
24. Statement: There’s at most one number b with the property that br = r for all real numbers
r.
Hint: First show that there exists an object with such a property and then show that if
objects A and B have the property then A=B.
Existence proof: When b=1, b*r=1*r = r for all real numbers r. Therefore, there exists at
least one such b.
Uniqueness proof: Suppose b1 and b2 are two real numbers with the above property for
all real numbers r. Then
(1) b1* r = r.
(2) b2* r = r.
Then b1*b2 = b2 by (1).
And b2*b1 = b1 by (2).
Then b2= b1*b2 = b2*b1 = b1. Hence b1 = b2. 